Question

A network of resistors is connected to a 16 V battery
with internal resistance of 1ohm. as shown. Find
(a) equivalent resistance of the network, (b) currents
in resistors R1, R3 and R4 (C) voltage drops across
R1 - R2, R3 and R4 - R5 resistors​

Answer 1

Answer:

Explanation:R1. R2. R3

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Lets first drive for series :

Consider three resistors R1,R2,R3 connected in series.

Let V1, V2, V3 be the current through them respectively.

We know, total current through them V = V1 + V2 + V3

i.e,

V = V1 + V2 + V3

IR = IR1+IR2+IR3. (since V= IR by ohm's law)

IR = I(R1+R2+R3)

R=R1+R2+R3

HENCE R=R1+R2+R3. (for series connection)

FOR PARALLEL:

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|____***____|

CONSIDER THREE RESISTORS R1,R2,R3 all covered

connected in parallel.

total I across resistors: I=I1+I2+I3

i.e,

I=I1+I2+I3

V/R= V/R1+V/R2+V/R3. (since V=IR and I=V/R).

V/R=V(1/R1+1/R2+1/R3)

1/R=1/R1+1/R2+1/R3

Thanks.

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